Question

Integration purely in cylindrical coordinates:

Answer 1

[Note: I'm not so interested with finding the solution to the problem in any way possible, I'm seeking specifically to solve it purely mathematically using cylindrical coordinates]

Answer 2

Find the electric field a distance z above the center of a circular loop of radius r that carries a uniform line charge, lambda.

For those not familiar with the physics behind it, the electric field at a point of a continuous charge distribution is, in general:

\[\vec{E}(\vec{r'})=\frac{1}{4\pi\epsilon_0}\cdot \int\limits \frac{\hat{r}}{r^2}dq\]Where r' is the position vector of the point of interest and r_hat is the vector from the infinitesimal charge to the point of interest.
\[dq=\lambda \cdot dl\]\[\vec{E}=\frac{1}{4 \pi \epsilon_0}\cdot \int\limits \frac{\vec{r}}{r^3}\lambda\ dl\]Pulling out constants:\[\vec{E}=\frac{1}{4 \pi \epsilon_0} \cdot \frac{\lambda}{r^3}\cdot \int\limits \vec{r}\ dl\]Based on the picture, we can define the r vector in cylindrical coordinates:\[\vec{r}=-r\hat{s}+z\hat{z}\]In addition, dl in cylindrical coordinates is:\[d\vec{l}=ds\hat{s}+sd \phi \hat{\phi}+dz\hat{z}\]Substituting in:\[\vec{E}=\alpha \cdot \int\limits (-r\hat{s}+z\hat{z})(ds\hat{s}+sd \phi \hat{\phi}+dz\hat{z})\]Alpha is all the stuff out front from before. ds and dz are 0, so those terms go to 0 and we get:\[\vec{E}=\alpha \cdot \int\limits_{\phi=0}^{2\pi} (-s^2 \hat{s}\hat{\phi}+zs\hat{z}\hat{\phi})d\phi\]Using r = s.

My question is whether there are any glaring flaws in my math so far? I'm not extremely comfortable with cylindrical coordinates, so I'm unsure.

Answer 3

The picture didn't show up:

Answer 4

Also, I'd like to solve the problem without making the obvious symmetry argument, whether conceptual or analytical.

Answer 5

If you're uncomfortable, simply do the computation in rectangular coordinates to check your answer since the electric field at a point has no bearing on what coordinate system you use.

You can always break a vector up into its components and add them back up later. So break them up into z components and then leave the x and y components behind.

If you don't want to do it with making a symmetry argument, just integrate this following integral with an arbitrary constant out front and you have your answer every time you want to make a symmetry argument, replace the arbitrary constant with whatever you have.

\[\int\limits_{0}^{2 \pi} [\cos(\theta) \hat x+\sin(\theta) \hat y] d \theta\]

Answer 6

I've done the problem in rectangular coordinates and I've done it while making the symmetry argument in cylindrical. I'm really just curious how, if it's possible, to do it in cylindrical without a switch to rectangular at any time and without the symmetry argument.

Answer 7

I think you are a bit confused with your choice of r as a variable.\[\vec{E}(\vec{r'})=\frac{1}{4\pi\epsilon_0}\cdot \int\limits \frac{\hat{r}}{r^2}dq\]here you are using \(r\) as the hypotenuse of the right triangle you drew, but in you other line\[\vec r=-r\hat s+z\hat z\]you seem to be using \(r\) as the radius of the circle...

Answer 8

in any case, when you multiply out\[\vec{E}=\alpha \cdot \int\limits (-r\hat{s}+z\hat{z})(ds\hat{s}+sd \phi \hat{\phi}+dz\hat{z})\] I don't see why you say ds is zero, but keep in mind that you can just foil it out, and that \(\hat s,\hat\theta,\hat z\) are orthogonal to each other, so \(\hat s\hat\theta=0\) and \(\hat s\hat s=1\).
Like I say though, I think you need to reevaluate your use of \(r\) first and keep it consistent.

Answer 9

I have a feeling to show cancellation without a symmetry argument you will need more than one integral, but I'm not sure.
I'd try to work it out with you, but I really should be reading up on my CS :P
Hope I gave a little direction, I'd like to see if you work it out

Answer 10

I was in a bit of a hurry last night when writing it, and some things didn't come through as clearly as I wanted them to.

Going to try and clarify everything for myself.


Using that notation and using the coordinate directions of s, phi, and z, I'll redefine some quantities so that they are more clear.

\[\vec{r}=-R\hat{s}+Z\hat{z}\]Also let r be\[r=|\vec{r}|\]Using these definitions to write the integral, we get:\[\vec{E}=\frac{1}{4\pi\epsilon_0}\cdot \int\limits\frac{\lambda}{r^3}\vec{r}\ dl=\frac{1}{4\pi\epsilon_0} \frac{\lambda}{r^3}\cdot \int\limits \vec{r}\ dl\]\[\vec{E}=\alpha \int\limits (-R\hat{s}+Z\hat{z})(ds\hat{s}+sd\phi\hat{\phi}+dz\hat{z})\]The problem I see with foiling out is:\[\vec{E}=\alpha \int\limits -R\ ds+\alpha \int\limits Z\ dz\]As we integrate along the ring, ds ranges from R to R and dz ranges from 0 to 0. Both of those integrals are 0, and this is where I run into trouble.